Question

Q2. Find four numbers in AP whose sum is 20 and the sum of whose squares is 120.
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Answer 1

Answer:

Let the numbers be (a−3d),(a−d),(a+d),(a+3d).

Then, Sum of numbers =20

⟹(a−3d)+(a−d)+(a+d)+(a+3d)=20⟹4a=20⟹a=5

It is given that, sum of the squares =120

⟹(a−3d)

2

+(a−d)

2

+(a+d)

2

+(a+3d)

2

=120

⟹4a

2

+20d

2

=120

⟹a

2

+5d

2

=30

⟹25+5d

2

=30

⟹5d

2

=5⟹d=±1

If d=1, the, the numbers are 2,4,6,8.

If d=−1, then the numbers are 8,6,4,2.

Thus, the numbers are 2,4,6,8 or 8,6,4,2.